What is the integral of x/(1+4x^4) ?

The integral of x/(1+4x^4) is 1/4 arctan(2x^2)+C

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integral of x/(1+4x^4)

\int \frac{x}{1 + 4 x ^{4}} d x

\frac{1}{4} arctan (2 x^{2}) + C

Solution steps

\int \frac{x}{1 + 4 x ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 ( 1 + 4 u ^{2} )} d u

= \int \frac{1}{2 ( 1 + 4 u ^{2} )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{1 + 4 u ^{2}} d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{2 ( 1 + v ^{2} )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int \frac{1}{1 + v ^{2}} d v

Use the common integral:

\int \frac{1}{1 + v ^{2}} d v = arctan (v)

= \frac{1}{2} \cdot \frac{1}{2} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{2} arctan (2 x^{2})

Simplify

\frac{1}{2} \cdot \frac{1}{2} arctan (2 x^{2}) : \frac{1}{4} arctan (2 x^{2})

= \frac{1}{4} arctan (2 x^{2})

Add a constant to the solution

= \frac{1}{4} arctan (2 x^{2}) + C

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What is the integral of x/(1+4x^4) ?
The integral of x/(1+4x^4) is 1/4 arctan(2x^2)+C